1.
The domain of a function f(x) is the range of values of x which is admitted by the function f.
For example, the domain of
f(x)=x²+2x-4
is (-∞,+∞), or all real numbers.
On the other hand, the domain of
f(x)=ln(x) is (0,∞), since the function ln(x) cannot accept non-positive values of x.
In this case, since we cannot hire a negative person, so the domain starts at ______ and goes up to _______.

The domain applies to the function _______=12L+29L^2-1.1L^3.

2.
For the maximum production, you will need to find dQ/dL and equate it to zero to find the maximum (or minimum).

1. So this is the answer I got 12L+29L^2-1.1L^3=0
L(12+29L-1.1L^2)=0
L=0, 26.77, -0.41
The minimum number of laborers would be 0.
The maximum number of laborers would be 26.77.
THe range would then be (0,26.77) because when I substitute a number >26.77 into the above equation it becomes a negative number. Am I on the right track or still way off?

2.Q'=12+58L-3.3L^2
12+58L-3.3L^2=0
L=17.78, -0.20
Q'(max)=17.78
To maximize the production they need 17.78 laborers

3. TO figure out how many laborers would be needed to maximize the revenue for producing cars?

How would I set that up?

4.Profit = price*production-7000L
L(-137.5L^2+3625L-5500)=0
L=0, 1.56, 24.9
The range would be 1.56 - 24.9. (I'm not sure if this is right).

How would I justify this answer with a graph? SHow the graphs of the 1st and 2nd derivatives?

Carry on with the problem.. To find maximum profit, not revenue, would I take the derivative of the profit function, which I have as
Profit=125*(12L+29L^2-1.1^3)-7000L. If I take the derivative of this(which I'm not sure I did right) my L=16.78 or .79. If I plug it back into the original profit function my max comes out to 278746.76. Does this look right?